剑指 Offer 55 – II. 平衡二叉树

输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

示例 1:

给定二叉树 [3,9,20,null,null,15,7]

返回 true 。

示例 2:

给定二叉树 [1,2,2,3,3,null,null,4,4]

返回 false 。

限制：

• 0 <= 树的结点个数 <= 10000

Python 解答：
1.笨方法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def depth(root):
            if not root:
                return 0
            else:
                left = depth(root.left)
                right = depth(root.right)
                return 1+max(left, right)
        flag = True
        def isHelper(root):
            if not root:
                return True
            else:
                isHelper(root.left)
                isHelper(root.right)
                if abs(depth(root.left)-depth(root.right)) > 1:
                    nonlocal flag
                    flag = False
        isHelper(root)
        return flag

2.求深度和判断合并

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        flag = True
        def depth(root):
            if not root:
                return 0
            else:
                left = depth(root.left)
                right = depth(root.right)
                if abs(left-right) > 1:
                    nonlocal flag
                    flag = False
                return 1+max(left, right)
        depth(root)
        return flag